Solving for X to Satisfy 2 Equations??

justinh8

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Aug 6, 2018
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Hi,

I have the following problem:

1. 27.5% = ((25+x)/(121+x))
2. 30.0% = ((43-x)/(121+x))

Im having issues solving for x that satisfy both the 27.5% and the 30%.

Please let me know of any structures I should be following.

Thanks!
 
Hi,

I have the following problem:

1. 27.5% = ((25+x)/(121+x))
2. 30.0% = ((43-x)/(121+x))

Im having issues solving for x that satisfy both the 27.5% and the 30%.

Please let me know of any structures I should be following.

Thanks!
How do you know that 'x' in equation 1 is same as that 2?
Why do you think that one value of "x" should satisfy both the equations - simultaneously?
 
Hi,

I have the following problem:

1. 27.5% = ((25+x)/(121+x))
0.275= (25+ x)/(121+ x)
Multiply both sides by 121+ x:
(0.275)(121+ x)= 33.275+ 0.275x= 25+ x
33.275- 25= x- 0.27x
8.275= 0.73x
x= 8.275/0.73= 11.335616438356164383561643835616

That is the only value of x that satisfies the equation.

2. 30.0% = ((43-x)/(121+x))
0.30= (43- x)/(121+ x)
0.30(121+ x)= 36.3+ 0.30x= 43- x
x+ 0.30x= 43- 36.3
1.30x= 5.7
x= 5.7/1.30= 4.3846153846153846153846153846154
That is the only value of x that satisfies the equation.

Im having issues solving for x that satisfy both the 27.5% and the 30%.

Please let me know of any structures I should be following.

Thanks!
That is because there is no value of x that satisfies both equations.
 
Hi,

I have the following problem:

1. 27.5% = ((25+x)/(121+x))
2. 30.0% = ((43-x)/(121+x))

Im having issues solving for x that satisfy both the 27.5% and the 30%.

Please let me know of any structures I should be following.

Thanks!
No such number exists.

\(\displaystyle \dfrac{25 + 11}{121 + 11} = \dfrac{36}{132} \approx 0.273 < 0.275.\)

\(\displaystyle \dfrac{25 + 12}{121 + 12} = \dfrac{37}{133} \approx 0.278 > 0.275.\)

So the answer to your first equation is between 11 and 12.

\(\displaystyle \dfrac{43 - 5}{121 + 5} = \dfrac{38}{126} \approx 0.302 > 0.3.\)

\(\displaystyle \dfrac{43 - 6}{121 + 6} = \dfrac{37}{127} \approx 29.1 < 0.3.\)

So the answer to your second equation is between 5 and 6.

But a single number cannot be less than 6 and more than 11.

EDIT: This is the same thought that Halls of Ivy expressed (though he made a slight arithmetic error, which is why I try to avoid using algebra because I always screw up the arithmetic.)
 
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A big cup of coffee for you while you stand in the corner?
I'm purty good at jumping to conclusions!
Anyhoo, I'll have company: both of Halls' solutions are wrong :cool:
 
I'm purty good at jumping to conclusions!
Anyhoo, I'll have company: both of Halls' solutions are wrong :cool:
Actually, it is a rather subtle mistake. I scratched my head over your response for quite a few minutes wondering whether I was wrong.

\(\displaystyle 0.275 = \dfrac{25 + x}{121 + x} \text { and } 0.3 = \dfrac{43 - y}{121 + y} \text { and } x = y \implies\)

\(\displaystyle 121 + x = 121 + y \implies \dfrac{43 - x}{0.3} =\dfrac{25 + x}{0.275} \implies\)

\(\displaystyle 11.825 - 0.275x = 7.5 + 0.3x \implies 0.575x = 4.325 \implies x = y = \dfrac{4325}{575} \approx 7.52.\)

So your equation can be solved. It just so happens that the solution does not work for either of the original equations.

The issue of course is that x is not equal to y. But since the problem is formulated using x in both equations, it is natural to assume that x stands for the same number in both equations. You really had me going there for a while.
 
and me too....

Denis ... go stand in the corner and don't Khan-fuse me like this ... ever again ...
 
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